Expectation of minimum discrete and continious random variable

Question

I have two random variables, one is $X$ which has $\mathbb{P}(X=2)=\mathbb{P}(X=4)=\mathbb{P}(X=6)=1/3$, and one, call it $Y$, which is $unif(3,5)$ (continious) distributed. I want to compute $\mathbb{E}(min(X,Y))$, how do I do this?

Answer 1

I'm guessing you're assuming that $X$ and $Y$ are independent. In which case, simply use the law of alternatives

$$E[min(X,Y)]=E[min(2,Y)|X=2]P[X=2]+E[min(4,Y)|X=4]P[X=4]+E[min(6,Y)|X=6]P[X=6]$$

$$=\frac{1}{3}(E[min(2,Y)]+E[min(4,Y)]+E[min(6,Y)])=\frac{1}{3}(E[2]+E[min(4,Y)]+E[Y])$$

Where I've used independence to get rid of the conditional expectations. All you have left is to compute what $E[min(4,Y)]$ is, again apply the law of alternatives using the events $\{Y\leq 4\}$ and $\{Y>4\}$.

Answer 2

Let $Z=\min(X,Y)$. Then, $$\mathbb{E}Z=\mathbb{E}(\mathbb{E}(Z|X))$$Now, $$\mathbb{E}(Z|X=k)=\mathbb{E}(\min (k,Y)|X=k),\ k=2,4,6$$ If $X\perp Y$, then $$\mathbb{E}(\min(k,Y)|X=k)=\mathbb{E}(\min(k,Y))=\int_{a}^k yf_{Y}dy+\int_{k}^b kf_Y(y) dy\\ \implies \mathbb{E}(\min(2,Y))=\int_{3}^5 2/2 dy=2$$ and similarl calculations for the others. So, $\mathbb{E}Z=\sum_{k}\mathbb{P}(X=k)\mathbb{E}(\min(Y,k))$

Answer 3

$$\mathbb{E}\min\left(X,Y\right)=\sum_{n=1}^{3}\mathbb{E}\left(\min\left(X,Y\right)\mid X=2n\right)P\left(X=2n\right)=$$$$\frac{1}{3}\sum_{n=1}^{3}\mathbb{E}\left(\min\left(2n,Y\right)\mid X=2n\right)$$

Here $\min\left(2,Y\right)=2$ and $\min\left(6,Y\right)=Y$ so we end up with:

$$\mathbb{E}\min\left(X,Y\right)=\frac{1}{3}\left[2+\mathbb{E}\left(\min\left(4,Y\right)\mid X=4\right)+\mathbb{E}\left(Y\mid X=6\right)\right]$$

If $X$ and $Y$ are independent then $$\mathbb{E}\left(Y\mid X=6\right)=\mathbb{E}Y$$ and $$\mathbb{E}\left(\min\left(4,Y\right)\mid X=4\right)=\mathbb{E}\min\left(4,Y\right)$$ The calculation of that I leave up to you.